Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, n\neq 0$. $\dfrac{{(k^{-2}n^{-2})^{5}}}{{(k^{-1}n^{3})^{-1}}}$
Explanation: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{-2}n^{-2})^{5} = (k^{-2})^{5}(n^{-2})^{5}}$ On the left, we have ${k^{-2}}$ to the exponent ${5}$ . Now ${-2 \times 5 = -10}$ , so ${(k^{-2})^{5} = k^{-10}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{-2}n^{-2})^{5}}}{{(k^{-1}n^{3})^{-1}}} = \dfrac{{k^{-10}n^{-10}}}{{kn^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-10}n^{-10}}}{{kn^{-3}}} = \dfrac{{k^{-10}}}{{k}} \cdot \dfrac{{n^{-10}}}{{n^{-3}}} = k^{{-10} - {1}} \cdot n^{{-10} - {(-3)}} = k^{-11}n^{-7}$